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Law Enforcement: Escaped Convicts Case studies showed that out of 10,351 convicts who escaped from U.S. prisons, only 7867 were recaptured (The Book of \(\mathrm{O} d d s\) by Shook and Shook, Signet). (a) Let \(p\) represent the proportion of all escaped convicts who will eventually be recaptured. Find a point estimate for \(p\). (b) Find a \(99 \%\) confidence interval for \(p .\) Give a brief statement of the meaning of the confidence interval. (c) Check Requirements Is use of the normal approximation to the binomial justified in this problem? Explain.

Short Answer

Expert verified
(a) \( \hat{p} \approx 0.7601 \); (b) \( (0.7489, 0.7713) \); (c) Requirements satisfied.

Step by step solution

01

Point Estimate for Proportion

To find a point estimate for the proportion \( p \), use the formula \( \hat{p} = \frac{x}{n} \), where \( x \) is the number of recaptured convicts and \( n \) is the total number of escaped convicts. In this case, \( x = 7867 \) and \( n = 10351 \). Therefore, \( \hat{p} = \frac{7867}{10351} \approx 0.7601 \). This is the point estimate for \( p \).
02

Find Confidence Interval

To find a \( 99\% \) confidence interval for \( p \), use the formula \( \hat{p} \pm z \times \sqrt{\frac{\hat{p}(1-\hat{p})}{n}} \), where \( z \) is the critical value for a \( 99\% \) confidence level, which is approximately 2.576. First, calculate the standard error (SE): \( SE = \sqrt{\frac{0.7601(1-0.7601)}{10351}} \approx 0.0043 \). Then, find the interval: \[ 0.7601 \pm 2.576 \times 0.0043 = (0.7489, 0.7713) \].
03

Interpret Confidence Interval

The \( 99\% \) confidence interval \((0.7489, 0.7713)\) indicates that we are \( 99\% \) confident that the true proportion of all escaped convicts who will eventually be recaptured lies between \( 74.89\% \) and \( 77.13\% \). This means if we were to take many samples of escaped convicts, \( 99\% \) of the time the proportion would fall within this interval.
04

Check Requirements for Normal Approximation

The normal approximation to the binomial distribution is justified if both \( np \) and \( n(1-p) \) are greater than 5. We check this for our problem: \( 10351 \times 0.7601 = 7867.28 \) and \( 10351 \times (1-0.7601) = 2483.72 \). Both values are significantly greater than 5, so the normal approximation is justified.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Confidence Interval
A confidence interval is a range of values that is used to estimate the true value of a population parameter. In our exercise on escaped convicts, we used a 99% confidence interval to estimate the proportion of escapees recaptured, denoted as \( p \). Calculating a confidence interval involves finding a range around the point estimate that we can be fairly sure contains the true parameter value.
To find a confidence interval for the proportion of convicts recaptured, you need:
  • A point estimate of the proportion, which we get from the sample data.
  • A critical value, which depends on the desired level of confidence (for 99%, this value is approximately 2.576).
  • The standard error of the proportion, which accounts for the sample size and variability.
In this context, the 99% confidence interval is \((0.7489, 0.7713)\). This suggests that we are 99% confident that the actual proportion of escaped prisoners who are recaptured lies between 74.89% and 77.13%. This high confidence level implies that if we perform the same experiment many times, this range would contain the true proportion 99% of the time.
Normal Approximation
Normal approximation is a technique used to approximate the binomial distribution using the normal distribution. This is especially useful when dealing with large sample sizes, as it simplifies calculations. The normal distribution is a continuous probability distribution, characterized by its bell shape.In statistical inquiries like our escaped convicts example, normal approximation to the binomial distribution is justified if both \( np \) and \( n(1-p) \) are greater than 5. This ensures that the sample size is sufficiently large such that the binomial distribution can be closely mimicked by the normal distribution. Indeed, for our case where \( n = 10,351 \), \( np = 7867.28 \) and \( n(1-p) = 2483.72 \), the values are well beyond 5. Therefore, we can use normal approximation for our calculations.
Binomial Distribution
The binomial distribution is a discrete probability distribution that arises in scenarios where there are only two possible outcomes, often referred to as "success" and "failure". It models the number of successes in a fixed number of trials.In the case of our exercise, the scenario can be modeled using a binomial distribution where each convict escape attempt is considered a trial. Success is defined as the recapture of an escaped convict. If we want to compute probabilities concerning the number of successes (recaptured convicts), we rely on this distribution. The binomial distribution is driven by two parameters: \( n \), the number of fixed trials (here, the number of escapees), and \( p \), the probability of success on each trial (probability of recapture).The nature of the binomial distribution makes it suitable for applications involving binary outcomes like detection and recovery of escaped convicts.
Point Estimate
A point estimate is a single value estimate for a population parameter. In statistics, this is often used as a summary measure to represent a characteristic of the population. The most straightforward method for estimating a population proportion \( p \) is to use the sample proportion, denoted as \( \hat{p} \).In our problem, the point estimate for the proportion of escaped convicts who were recaptured is calculated using data from a survey or study. By taking the number of captured convicts and dividing it by the total number of escapes, we obtain \( \hat{p} = \frac{7867}{10351} \approx 0.7601 \).This result tells us that, based on the sample data, approximately 76.01% of all escaped convicts are expected to be recaptured. The point estimate \( \hat{p} \) provides a direct insight into our specific case study by drawing conclusions about the entire group from the study sample.

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Most popular questions from this chapter

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